To solve the differential equation:

\[
\frac{dy}{dx} = y \cdot e^x
\]

with the initial condition \( y(0) = e^6 \), we can use the method of separation of variables.

### Step 1: Separate Variables
Rewrite the equation by dividing both sides by \( y \) and multiplying by \( dx \):

\[
\frac{1}{y} \, dy = e^x \, dx
\]

### Step 2: Integrate Both Sides
Integrate both sides with respect to their respective variables:

\[
\int \frac{1}{y} \, dy = \int e^x \, dx
\]

The left side integrates to \( \ln |y| \), and the right side integrates to \( e^x \):

\[
\ln |y| = e^x + C
\]

where \( C \) is the constant of integration.

### Step 3: Solve for \( y \)
Exponentiate both sides to solve for \( y \):

\[
y = e^{e^x + C} = e^C \cdot e^{e^x}
\]

Let \( e^C = K \), where \( K \) is a constant. Then we have:

\[
y = K \cdot e^{e^x}
\]

### Step 4: Apply the Initial Condition
Use the initial condition \( y(0) = e^6 \) to find \( K \):

\[
y(0) = K \cdot e^{e^0} = K \cdot e^1 = K \cdot e
\]

Since \( y(0) = e^6 \), we get:

\[
K \cdot e = e^6
\]

Dividing by \( e \), we find \( K = e^5 \).

### Final Solution
Substitute \( K = e^5 \) back into the solution:

\[
y = e^5 \cdot e^{e^x} = e^{5 + e^x}
\]

So, the solution to the differential equation is:

\[
y = e^{5 + e^x}
\]

Would you like further details on any step? 

Here are 5 questions related to this topic:

1. What are other methods to solve first-order differential equations?
2. How does the initial condition influence the constant of integration?
3. What if \( y(0) \) were a different value; how would that change the solution?
4. Can you use this method of separation of variables for all differential equations?
5. How do you verify that this solution satisfies the original differential equation?

**Tip:** When solving differential equations, always remember to check if the solution meets the initial conditions by plugging them back into your general solution.

Learn how to solve the first-order differential equation dy/dx = y * e^x using separation of variables. This solution includes step-by-step integration, applying initial conditions, and finding the final expression y = e^{5 + e^x}. Suitable for advanced high school or undergraduate math students.

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